$\begin{array}{1 1}(A)\;180\\(B)\;190\\(C)\;200\\(D)\;210\end{array} $

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- $n!=n(n-1)(n-2)(n-3)...(3)(2)(1)$

Let us draw line 1 by 1

$1^{st}$ line =0 points

$2^{nd}$ line =new 1 point

$3^{rd}$ line =new 2 points+old 1 point

$4^{th}$ line =new 3 points +old 2+1 points

$n^{th}$ line =(n-1)points +(n-2).....(3)(2)(1)

$\therefore S=1+2+3.....(n-1)$

$S=(n-1)\times \large\frac{n}{2}$

$\Rightarrow 19\times \large\frac{20}{2}$

$\Rightarrow 19\times 10$

$\Rightarrow 190$

Hence (B) is the correct answer.

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